Answer
Verified
449.4k+ views
Hint: Incase of longitudinal waves take an example of the sound wave and we have a formula to find out the speed of sound wave in particular medium having certain elastic modulus(E). Similarly we have a formula to find out what will be the speed of the transverse wave in the wire having tension(T) and linear mass density ($\mu $). Then we equate the speed of the longitudinal wave to ten times the speed of the transverse wave to find out stress in the wire.
Formula used:
$\eqalign{
& {S_L} = \sqrt {\dfrac{E}{\rho }} \cr
& {S_T} = \sqrt {\dfrac{T}{\mu }} \cr} $
$\eqalign{
& \dfrac{T}{A} = stress \cr
& \dfrac{{\Delta l}}{l} = strain \cr
& \dfrac{{stress}}{{strain}} = E \cr} $
Complete answer:
Let the speed of longitudinal wave in the stretched wire be SL and speed of the transverse wave in the stretched wire be ST
We know that
$\eqalign{
& {S_L} = \sqrt {\dfrac{E}{\rho }} \cr
& {S_T} = \sqrt {\dfrac{T}{\mu }} \cr} $
Where
E = elastic modulus or young’s modulus of elasticity
ρ = volumetric mass density of wire
T = tension in the wire
$\mu $= linear mass density of wire
Now
$\eqalign{
& \dfrac{T}{A} = stress \cr
& \dfrac{{\Delta l}}{l} = strain \cr
& \dfrac{{stress}}{{strain}} = E \cr} $
Where
A = area of cross section of wire
l = initial length of the wire
$\Delta l$ = variation in length of wire due to tension applied i.e final length – initial length
STRESS/STRAIN = E
STRESS = STRAIN $\times$ E
$\dfrac{T}{A} = \dfrac{{\Delta l}}{l} \times E$ … eq 1
$\eqalign{
& \rho = \dfrac{\mu }{A} \cr
& \mu = \rho \times A \cr
& {S_T} = \sqrt {\dfrac{T}{\mu }} = \sqrt {\dfrac{T}{{\rho \times A}}} = \sqrt {\dfrac{{E \times \dfrac{{\Delta l}}{l}}}{\rho }} = {S_L} \times \sqrt {\dfrac{{\Delta l}}{l}} \cr} $
Since
${S_L} = \sqrt {\dfrac{E}{\rho }} $
It is given in the question that ratio of longitudinal wave speed and transverse wave speed is 10
By equating that and from equation 1 we get
$\eqalign{
& {S_L} = 10 \times {S_T} \cr
& \Rightarrow \dfrac{{{S_T}}}{{{S_L}}} = \dfrac{1}{{10}} \cr
& \Rightarrow \dfrac{{{S_T}}}{{{S_L}}} = \sqrt {\dfrac{{\Delta l}}{l}} \cr
& \Rightarrow \sqrt {\dfrac{{\Delta l}}{l}} = \dfrac{1}{{10}} \cr
& \Rightarrow \dfrac{{\Delta l}}{l} = \dfrac{1}{{100}} \cr
& \Rightarrow \dfrac{T}{A} = \dfrac{1}{{100}} \times E \cr
& \Rightarrow STRESS = \dfrac{E}{{100}} \cr} $
So, the correct answer is “Option B”.
Note:
The velocities which we consider over here are velocities of waves in particular medium but not velocities of individual particles oscillating in that wave. Velocity of waves is constant as long as material density and source frequency remain constant but particle velocity at every position of wave is different. At maximum displacement position of wave, particle velocity there is minimum and at minimum displacement position particle velocity is maximum.
Formula used:
$\eqalign{
& {S_L} = \sqrt {\dfrac{E}{\rho }} \cr
& {S_T} = \sqrt {\dfrac{T}{\mu }} \cr} $
$\eqalign{
& \dfrac{T}{A} = stress \cr
& \dfrac{{\Delta l}}{l} = strain \cr
& \dfrac{{stress}}{{strain}} = E \cr} $
Complete answer:
Let the speed of longitudinal wave in the stretched wire be SL and speed of the transverse wave in the stretched wire be ST
We know that
$\eqalign{
& {S_L} = \sqrt {\dfrac{E}{\rho }} \cr
& {S_T} = \sqrt {\dfrac{T}{\mu }} \cr} $
Where
E = elastic modulus or young’s modulus of elasticity
ρ = volumetric mass density of wire
T = tension in the wire
$\mu $= linear mass density of wire
Now
$\eqalign{
& \dfrac{T}{A} = stress \cr
& \dfrac{{\Delta l}}{l} = strain \cr
& \dfrac{{stress}}{{strain}} = E \cr} $
Where
A = area of cross section of wire
l = initial length of the wire
$\Delta l$ = variation in length of wire due to tension applied i.e final length – initial length
STRESS/STRAIN = E
STRESS = STRAIN $\times$ E
$\dfrac{T}{A} = \dfrac{{\Delta l}}{l} \times E$ … eq 1
$\eqalign{
& \rho = \dfrac{\mu }{A} \cr
& \mu = \rho \times A \cr
& {S_T} = \sqrt {\dfrac{T}{\mu }} = \sqrt {\dfrac{T}{{\rho \times A}}} = \sqrt {\dfrac{{E \times \dfrac{{\Delta l}}{l}}}{\rho }} = {S_L} \times \sqrt {\dfrac{{\Delta l}}{l}} \cr} $
Since
${S_L} = \sqrt {\dfrac{E}{\rho }} $
It is given in the question that ratio of longitudinal wave speed and transverse wave speed is 10
By equating that and from equation 1 we get
$\eqalign{
& {S_L} = 10 \times {S_T} \cr
& \Rightarrow \dfrac{{{S_T}}}{{{S_L}}} = \dfrac{1}{{10}} \cr
& \Rightarrow \dfrac{{{S_T}}}{{{S_L}}} = \sqrt {\dfrac{{\Delta l}}{l}} \cr
& \Rightarrow \sqrt {\dfrac{{\Delta l}}{l}} = \dfrac{1}{{10}} \cr
& \Rightarrow \dfrac{{\Delta l}}{l} = \dfrac{1}{{100}} \cr
& \Rightarrow \dfrac{T}{A} = \dfrac{1}{{100}} \times E \cr
& \Rightarrow STRESS = \dfrac{E}{{100}} \cr} $
So, the correct answer is “Option B”.
Note:
The velocities which we consider over here are velocities of waves in particular medium but not velocities of individual particles oscillating in that wave. Velocity of waves is constant as long as material density and source frequency remain constant but particle velocity at every position of wave is different. At maximum displacement position of wave, particle velocity there is minimum and at minimum displacement position particle velocity is maximum.
Recently Updated Pages
Who among the following was the religious guru of class 7 social science CBSE
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Trending doubts
Derive an expression for drift velocity of free electrons class 12 physics CBSE
Which are the Top 10 Largest Countries of the World?
Write down 5 differences between Ntype and Ptype s class 11 physics CBSE
The energy of a charged conductor is given by the expression class 12 physics CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Derive an expression for electric field intensity due class 12 physics CBSE
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Derive an expression for electric potential at point class 12 physics CBSE